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(i) (7, -2), (5, 1), (3, -k)

(ii) (8, 1), (k, -4), (2, -5)

Answer :

We know that for collinear points area of triangle = 0 ,i.e.,

\( 0 \ = \frac{1}{2}[x_1(y_2 - y_3) \ + \ x_2(y_3 - y_1) \ + \ x_3(y_1 - y_2)] \)

(i) \( 7(1-k) + 5(k+2) + 3(-2-1) \ = \ 0 \)

=> \( 7 - 7k + 5k + 10 - 9 \ = \ 0 \)

=> \( 2k \ = \ 8 \)

=> \( k \ = \ 4 \)

(ii) \( 8(-4+5) + k(-5-1) + 2(1+4) \ = \ 0 \)

=> \( 8 - 6k + 10 \ = \ 0 \)

=> \( 6k \ = \ 18 \)

=> \( k \ = \ 3 \)

- Find the area of the triangle whose vertices are: (i) (2, 3), (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)
- Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
- Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
- You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for \(\triangle \) ABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).

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